Problem: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $19.4$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $16.1$ years.
Solution: $19.4$ $16.1$ $22.7$ $12.8$ $26$ $9.5$ $29.3$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $19.4$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $16.1$ years and one standard deviation above the mean is $22.7$ years. Two standard deviations below the mean is $12.8$ years and two standard deviations above the mean is $26$ years. Three standard deviations below the mean is $9.5$ years and three standard deviations above the mean is $29.3$ years. We are interested in the probability of a tiger living less than $16.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $16.1$ years and the other half $({16\%})$ will live longer than $22.7$ years. The probability of a particular tiger living less than $16.1$ years is ${16\%}$.